A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :

Question

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are : (A) 10 cm, 12 cm (B) 12 cm, 14 cm (C) 16 cm, 18 cm (D) 18 cm, 20 cm

Tardigrade Admin · Accepted Answer

For a convergent doublet of separated lens, we have

\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} - \frac{d}{f _{1} f _{2}}

where

d

is separation between two lens,

f_{1}

and

f_{2}

are focal lengths of component lenses,

f

is resultant focal length. Therefore, Eq.

(1)

becomes

\frac{1}{10} = \frac{1}{f _{1}} + \frac{1}{f _{2}} - \frac{2}{f _{1} f _{2}} \Rightarrow \frac{1}{10} = (\frac{f _{2} + f _{1} - 2}{f _{1} f _{2}})

\Rightarrow f f_{2} = 10 f_{2} + 10 f_{1} - 20

\Rightarrow 10 f_{1} + 10 f_{2} - f f_{2} = + 20

For

f = 18 c m

and

f_{2} = 20 c m

, the above equation satisfies.

Q. A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10cm. The separation between the two lenses is 2cm. The focal lengths of the component lenses are :

10 cm, 12 cm

12 cm, 14 cm

16 cm, 18 cm

18 cm, 20 cm

Q. A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of $10 c m$ . The separation between the two lenses is $2 c m$ . The focal lengths of the component lenses are :